MHB How can we prove that a Pythagorean triple is primitive?

  • Thread starter Thread starter Poirot1
  • Start date Start date
  • Tags Tags
    Primitive
AI Thread Summary
To prove that a Pythagorean triple (x, y, z) = (a^2 - b^2, 2ab, a^2 + b^2) is primitive, it is essential to establish that gcd(x, y, z) = 1. This can be shown by ensuring that the integers a and b are coprime (gcd(a, b) = 1) and of opposite parity, meaning one is even and the other is odd. A contradiction argument can also be employed, assuming a common factor d greater than 1, leading to a prime factorization. The discussion emphasizes the importance of these conditions in confirming the primitiveness of the triple. Understanding these principles is crucial for mathematical proofs involving Pythagorean triples.
Poirot1
Messages
243
Reaction score
0
I know that (a^2-b^2,2ab,a^2+b^2) is pythagorean triple. How to show it is primitive? i.e
gcd(x,y,z)=1
 
Mathematics news on Phys.org
Re: primitive pythagorean triple

Hello, Poirot!

I know that (x,y,z) = (a2 - b2, 2ab, a2 + b2) is a Pythagorean triple.

How to show it is primitive? .i.e. gcd(x,y,z) = 1
I'm not sure how we "show" it, but here is a fact.

For a primitive Pythagorean triple, a and b must be of opposite parity.
. . (One must be even, the other must be odd.)
 
Re: primitive pythagorean triple

Poirot said:
I know that (a^2-b^2,2ab,a^2+b^2) is pythagorean triple. How to show it is primitive? i.e
gcd(x,y,z)=1
The conditions for the triple to be primitive are that gcd(a,b)=1 and a, b are of opposite parity. See Pythagorean triple - Wikipedia, the free encyclopedia.
 
Re: primitive pythagorean triple

I'm pretty sure a contradiction argument is expedient.

assume x,y,z have a common factor d not equal to 1.
Since everything can be factorized into primes can I assume d is prime?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top